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Old Jun 16th 2017, 06:00 PM   #1
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Time dilation problem

Hi all,

I have attempted the attached problem;

The issue I has is I've seen it done 2 ways;

1. Calculate the proper time for the period of the clock then convert that time to the time that passes in the observers reference frame... This gives 42 ns.
I believe this method to be correct.

2. The second method I've seen used (which I believe must be incorrect) is to calculate the length the observer would see for the clock using 5m as the proper length. (comes out to be 4m) Then use t=d/c to calculate the time the observer would see light take to complete the distance. this results in a different answer.

My question is why does the second method (I'm assuming that's the incorrect approach) not work?

I'm convinced there's something I'm overlooking here I just need some help figuring out what?

Many thanks in advance.
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Old Jun 17th 2017, 03:51 PM   #2
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It should be theoretically possible to apply #2 method (although its more work) and get the same answer. Will look into it and see what I come up with.
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Old Jun 17th 2017, 06:41 PM   #3
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Ok, I figured it out and got same result for #1 and #2.

Did you remember to take into account length contraction of the length of the light clock? (It will end up being less than 5m for the stationary observer). Once I compensated for that both #1 and #2 agreed in result.
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Old Jun 18th 2017, 05:25 AM   #4
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I still cant get it! :-(

When i calculate length contraction i get L=4 then i double that as light must travel twice the length, then when i divide by c i get 26 ns. What am i doing wrong?

Thanks for the help too its really bugging me
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Old Jun 18th 2017, 12:11 PM   #5
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How are you calculating the period of the clock in #2?

Remember, for the reference frame of the stationary observer the spaceship and the light clock inside of it are moving so that a light pulse leaving the left hand edge and travelling to the right edge of the clock won't reach it in L/(gamma*c) because the light clock will also have moved a certain distance, because the entire spaceship has moved, in that time
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Old Jun 18th 2017, 03:05 PM   #6
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Ok thats the part i cant get my head around, i figured that it would travel further in the drection of motion but then a smaller distance back against motion as the ship is moving and i thought that this would amount to having travlled L x 2 for the period?
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Old Jun 18th 2017, 06:11 PM   #7
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From the point of view of the stationary observer the light pulse left to right distance travelled is not the same as the right to left distance travelled. The whole light clock apparatus is moving at a speed of 0.6c. When the light pulse travels a length of 5m (or 5m/1.25 for the stationary observer) the light clock has moved from its original position. To find the time taken from left to right you can solve the simultaneous linear equations x = c*t (eqn of position of light pulse) and x = 5metres/1.25+ 0.6*c*t (eqn of position of right hand edge of light clock). On the reverse trip it will be similar but you will have a sign change on the second eqn. The two pulse travel times will not be equal for the stationary observer. The first trip will be longer. Does that make more sense? Can you follow through from there?
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Last edited by kiwiheretic; Jun 18th 2017 at 06:15 PM.
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